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That depends on the value of the angle, theta. csc is short for "cosecans", and is the reciprocal of the sine. That is, csc theta = 1 / sin theta.

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โˆ™ 2012-10-30 10:57:14
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A polynomial of degree zero is a constant term

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Q: What is csc theta?
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Related questions

What is cos theta multiplied by csc theta?

It is cotangent(theta).


How do i simplify csc theta divided by sec theta?

By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.


How do you get the csc theta given tan theta in quadrant 1?

If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)


Express csc theta in terms of cot theta theta is in quadrant 3?

It is -sqrt(1 + cot^2 theta)


Verify that Cos theta cot theta plus sin theta equals csc theta?

It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)


How do you simplify csc theta -cot theta cos theta?

For a start, try converting everything to sines and cosines.


How do you simplify cos theta times csc theta divided by tan theta?

'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2


What is a trig function?

sin theta and csc theta are reciprocal functions because sin = y/r and csc = r/y you use the same 2 sides of a triangle, but you use the reciprocal.


How do you simplify sin theta times csc theta divided by tan theta?

Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).


What is a reciprocal trig function?

sin theta and csc theta are reciprocal functions because sin = y/r and csc = r/y you use the same 2 sides of a triangle, but you use the reciprocal.


How do you simplify csc theta cot theta cos theta?

cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)


What is the exact value of cos theta if csc theta -4 with theta in quadrant III?

csc θ = 1/sin θ → sin θ = -1/4 cos² θ + sin² θ = 1 → cos θ = ± √(1 - sin² θ) = ± √(1 - ¼²) = ± √(1- 1/16) = ± √(15/16) = ± (√15)/4 In Quadrant III both cos and sin are negative → cos θ= -(√15)/4

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